Optimizing curvature of $\frac{\sin(x)}x$

Question

At what point along the curve does the graph of $y=\frac{\sin(x)}x$ have the smallest curvature?

I've tried computing this with a bunch of derivatives but I don't think it's meant to be solved that way

Answer 1

I am not sure that we could have an explicit solution.

Let $$y=\frac{\sin(x)}x \implies y'=\frac{x \cos (x)-\sin (x)}{x^2}\implies y''=-\frac{\left(x^2-2\right) \sin (x)+2 x \cos (x)}{x^3}$$ Minimizing $$\kappa = \frac{|y''|}{\left(1+(y')^2\right)^\frac32}$$ is the same as minimizing $\kappa ^2$ and $$\frac {d\kappa ^2}{dx}=\frac{2 y'' \left(y''' \left(1+(y')^2\right)-3 y' (y'')^2\right)}{\left(1+(y')^2\right)^4}$$ So, one solution corresponds to $y''=0$ (in fact, if you look here), you will notice the approximation $\kappa\approx y''$) and then we need to solve for $x$ $$\left(x^2-2\right) \sin (x)+2 x \cos (x)=0$$ that is to say to find the zero of function $$f(x)=\left(x^2-2\right) \sin (x)+2 x \cos (x)$$ If you graph this function or use inspection, you will notice that there is a root close to $x_0=\frac{2 \pi }{3}$. So, let us use Newton method to get the following iterates $$x_{n+1}=\frac{\left(x_n^2-2\right) (x_n-\tan (x_n))}{x_n^2}$$ $$\left( \begin{array}{cc} n & x_n \\ 0 & 2.094395102 \\ 1 & 2.081795781 \\ 2 & 2.081576044 \\ 3 & 2.081575978 \end{array} \right)$$ which is the solution for ten significant figures.

You could skip the Newton part building a series expansion close to $x_0$. This would give $$y''=\left(\sqrt{3}+\frac{2 \pi }{3}-\frac{2 \pi ^2}{3 \sqrt{3}}\right)+\frac{2}{9} \pi ^2 \left(x-\frac{2 \pi }{3}\right)+O\left(\left(x-\frac{2 \pi }{3}\right)^2\right)$$ and, ignoring the higher order terms, this would give the estimate $$x=\frac{2 \pi }{3}-\left(\frac{9 \sqrt{3}}{2 \pi ^2}+\frac{3}{\pi }-\sqrt{3} \right)\approx 2.08180$$

We could have a still better approximation if, instead of using Taylor series, we build the $[1,1]$ Padé approximant of $f(x)$ around $x=\frac{2 \pi }{3}$. This would lead to $$x=\frac{2 \pi }{3}+\frac{2 \pi \left(9 \sqrt{3}+6 \pi -2 \sqrt{3} \pi ^2\right)}{27 \sqrt{3}+45 \pi -10 \pi ^3}\approx 2.08158$$